# The Unapologetic Mathematician

## The Automorphism Group of a Root System

Finally, we’re able to determine the automorphism group of our root systems. That is, given an object in the category of root systems, the morphisms from that root system back to itself (as usual) form a group, and it’s interesting to study the structure of this group.

First of all, right when we first talked about the category of root systems, we saw that the Weyl group $\mathcal{W}$ of $\Phi$ is a normal subgroup of $\mathrm{Aut}(\Phi)$. This will give us most of the structure we need, but there may be automorphisms of $\Phi$ that don’t come from actions of the Weyl group.

So fix a base $\Delta$ of $\Phi$, and consider the collection $\Gamma$ of automorphisms which send $\Delta$ back to itself. We’ve shown that the action of $\mathcal{W}$ on bases of $\Phi$ is simply transitive, which means that if $\tau\in\Gamma$ comes from the Weyl group, then $\tau$ can only be the identity transformation. That is, $\Gamma\cap\mathcal{W}=\{1\}$ as subgroups of $\mathrm{Aut}(\Phi)$.

On the other hand, given an arbitrary automorphism $\tau\in\mathrm{Aut}(\Phi)$, it sends $\Delta$ to some other base $\Delta'$. We can find a $\sigma\in\mathcal{W}$ sending $\Delta'$ back to $\Delta$. And so $\sigma\tau\in\Gamma$; it’s an automorphism sending $\Delta$ to itself. That is, $\tau\in\mathcal{W}\Gamma$; any automorphism can be written (not necessarily uniquely) as the composition of one from $\Gamma$ and one from $\mathcal{W}$. Therefore we can write the automorphism group as the semidirect product:

$\displaystyle\mathrm{Aut}(\Phi)=\Gamma\ltimes\mathcal{W}$

All that remains, then, is to determine the structure of $\Gamma$. But each $\tau\in\Gamma$ shuffles around the roots in $\Delta$, and these roots correspond to the vertices of the Dynkin diagram of the root system. And for $\tau$ to be an automorphism of $\Phi$, it must preserve the Cartan integers, and thus the numbers of edges between any pair of vertices in the Dynkin diagram. That is, $\Gamma$ must be a transformation of the Dynkin diagram of $\Phi$ back to itself, and the reverse is also true.

So we can determine $\Gamma$ just by looking at the Dynkin diagram! Let’s see what this looks like for the connected diagrams in the classification theorem, since disconnected diagrams just add transformations that shuffle isomorphic pieces.

Any diagram with a multiple edge — $G_2$, $F_4$, and the $B_n$ and $C_n$ series — has only the trivial symmetry. Indeed, the multiple edge has a direction, and it must be sent back to itself with the same direction. It’s easy to see that this specifies where every other part of the diagram must go.

The diagram $A_1$ is a single vertex, and has no nontrivial symmetries either. But the diagram $A_n$ for $n\geq2$ can be flipped end-over-end. We thus find that $\Gamma=\mathbb{Z}_2$ for all these diagrams. The diagram $E_6$ can also be flipped end-over-end, leaving the one “side” vertex fixed, and we again find $\Gamma=\mathbb{Z}_2$, but $E_7$ and $E_8$ have no nontrivial symmetries.

There is a symmetry of the $D_n$ diagram that swaps the two “tails”, so $\Gamma=\mathbb{Z}_2$ for $n\geq5$. For $n=4$, something entirely more interesting happens. Now the “body” of the diagram also has length ${1}$, and we can shuffle it around just like the “tails”. And so for $D_4$ we find $\Gamma=S_3$ — the group of permutations of these three vertices. This “triality” shows up in all sorts of interesting applications that connect back to Dynkin diagrams and root systems.